CUDA 中的递归返回非法内存访问

我正在编写一个数值积分程序，该程序实现了具有自适应步长的梯形规则。在不赘述太多细节的情况下，该算法使用递归来计算具有指定相对公差的给定区间内编码数学函数的积分。 我简化了发布代码，但保留了所有要点，因此某些部分可能看起来不必要或过于复杂。在这里：

#include "cuda_runtime.h"
#include "device_launch_parameters.h"
#include <cmath>
#include <iostream>
#include <iomanip>
class Integral {
public:
double value;       // the value of the integral
__device__ __host__ Integral() : value(0) {};
__device__ __host__ Integral& operator+=(Integral &I);
};
__device__ Integral trapezoid(double a, double b, double tolerance, double fa, double fb);
__device__ __host__ double f(double x); // the integrand function
const int BLOCKS = 1;
const int THREADS = 1;
__global__ void controller(Integral *blockIntegrals, double a, double b, double tolerance) {
extern __shared__ Integral threadIntegrals[]; // an array of thread-local integrals
double fa = f(a), fb = f(b);
threadIntegrals[threadIdx.x] += trapezoid(a, b, tolerance, fa, fb);
blockIntegrals[blockIdx.x] += threadIntegrals[0];
}
int main() {
// *************** Input parameters ***************
double a = 1, b = 800;  // integration bounds
double tolerance = 1e-7;
// ************************************************
cudaError cudaStatus;
Integral blockIntegrals[BLOCKS]; // an array of total integrals computed by each block
Integral *devBlockIntegrals;
cudaStatus = cudaMalloc((void**)&devBlockIntegrals, BLOCKS * sizeof(Integral));
if (cudaStatus != cudaSuccess)
std::cout << "cudaMalloc failed!n";
double estimate = 0; // a rough 10-point estimate of the whole integral
double h = (b - a) / 10;
for (int i = 0; i < 10; i++)
estimate += f(a + i*h);
estimate *= h;
tolerance *= estimate; // compute relative tolerance
controller<<<BLOCKS, THREADS, THREADS*sizeof(Integral)>>>(devBlockIntegrals, a, b, tolerance);
cudaStatus = cudaGetLastError();
if (cudaStatus != cudaSuccess)
std::cout << "addKernel launch failed: " << cudaGetErrorString(cudaStatus) << "n";
cudaStatus = cudaMemcpy(blockIntegrals, devBlockIntegrals, BLOCKS * sizeof(Integral), cudaMemcpyDeviceToHost);
if (cudaStatus != cudaSuccess)
std::cout << "cudaMemcpy failed: " << cudaGetErrorString(cudaStatus) << "n";
Integral result; // final result
for (int i = 0; i < BLOCKS; i++) // final reduction that sums the results of all blocks
result += blockIntegrals[i];
std::cout << "Integral = " << std::setprecision(15) << result.value;
cudaFree(devBlockIntegrals);
getchar();
return 0;
}
__device__  double f(double x) {
return log(x);
}
__device__ Integral trapezoid(double a, double b, double tolerance, double fa, double fb) {
double h = b - a;               // compute the new step
double I1 = h*(fa + fb) / 2;    // compute the first integral
double m = (a + b) / 2;         // find the middle point
double fm = f(m);                       // function value at the middle point
h = h / 2;                              // make step two times smaller
double I2 = h*(0.5*fa + fm + 0.5*fb);   // compute the second integral
Integral I;
if (abs(I2 - I1) <= tolerance) {    // if tolerance is satisfied
I.value = I2;
}
else {  // if tolerance is not satisfied
if (tolerance > 1e-15) // check that we are not requiring too high precision
tolerance /= 2; // request higher precision in every half
I += trapezoid(a, m, tolerance, fa, fm);    // integrate the first half [a m]
I += trapezoid(m, b, tolerance, fm, fb);    // integrate the second half [m b]
}
return I;
}
__device__ Integral& Integral::operator+=(Integral &I) {
this->value += I.value;
return *this;
}

为简单起见，我在这里只使用一个线程。 现在，如果我运行此代码，我会收到一条消息"cudaMemcpy 失败：遇到非法内存访问"。当我运行"cuda-memcheck"时，我收到此错误：

========= Invalid __local__ write of size 4
=========     at 0x00000b18 in C:/Users/User/Desktop/Integrator Stack/Integrator_GPU/kernel.cu:73:controller(Integral*, double, double, double)
=========     by thread (0,0,0) in block (0,0,0)
=========     Address 0x00fff8ac is out of bounds

它说问题出在 73 行，它只是

double m = (a + b) / 2;

难道此时我的内存不足？

如果我通过在main中将右边界从b = 800更改为b = 700来缩小积分间隔，则程序运行良好，并且给出了正确的结果。 为什么我在简单地创建新变量时收到非法内存访问错误？

另外，我有一个相同的CPU版本的该程序，并且它完美运行，因此计算算法很可能是正确的。